Basis Of A Three Dimensional Space

Go back to  'Vectors and 3-D Geometry'

In the preceeding discussion, we talked about the basis of a plane. We can easily extend that discussion to observe that any three non-coplanar vectors can form a basis of three dimensional space:

In other words, any vector \(\vec r\) in 3-D space can be expressed as a linear combination of three arbitrary non-coplanar vectors. From this, it also follows that for three non-coplanar vectors \(\vec a,\vec b,\vec c,\) if their linear combination is zero, i.e, if

\[\lambda \vec a + \mu \vec b + \gamma \vec c = \vec 0\quad\qquad\qquad\left( {{\text{where}}\,\,\lambda ,\mu ,\gamma  \in \mathbb{R}} \right)\]

then \(\lambda ,\mu \,\,{\text{and}}\,\,\gamma \) must all be zero. To prove this, assume the contrary. Then, we have

\[\vec a = \left( { - \frac{\mu }{\lambda }} \right)\vec b + \left( { - \frac{\gamma }{\lambda }} \right)\vec c\]

which means that \(\vec a\) can be written as the linear combination of  \(\vec b\,\,{\text{and}}\,\,\vec c\). However, this would make \(\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c\) coplanar, contradicting our initial supposition. Thus, \(\lambda ,\mu \,\,{\text{and}}\,\,\gamma \)  must be zero.

We finally come to what we mean by linearly independent and linearly dependent vectors.

Linearly independent vectors : A set of non-zero vectors \({\vec a_1},{\vec a_2},{\vec a_3}....,{\vec a_n}\)  is said to be linearly independent if

\[{\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ... + {\lambda _n}{\vec a_n} = \vec 0\]

\[implies\,\,\,\,{\lambda _1} = {\lambda _2} = .... = {\lambda _n} = 0\]

Thus, a linear combination of linearly independent vectors cannot be zero unless all the scalars used to form the linear combination are zero.

Linearly dependent  vectors: A set of non-zero vectors  \({\vec a_1},{\vec a_2},{\vec a_3},....,{\vec a_n}\)  is said to be linearly dependent if there exist scalars \({\lambda _1},{\lambda _2}....{\lambda _n},\) not all zero such that,

\[{\lambda _1}{\vec a_1} + {\lambda _2}{\vec a_2} + ..... + {\lambda _n}{\vec a_n} = \vec 0\]

For example, based on our previous discussions, we see that

(i) Two non-zero, non-collinear vectors are linearly independent.

(ii) Two collinear vectors are linearly dependent

(iii) Three non-zero, non-coplanar vectors are linearly independent.

(iv) Three coplanar vectors are linearly dependent

(v) Any four vectors in 3-D space are linearly dependent.

You are urged to prove for yourself all these assertions.

Example – 5

Let \({\vec a},\vec b\,\,{\text{and}}\,\,\vec c\) be non-coplanar vectors. Are the vectors \(2\vec a - \vec b + 3\vec c,\,\,\vec a + \vec b - 2\vec c\,\,{\text{and}}\,\,\vec a + \vec b - 3\vec c\)  coplanar or non-coplanar?

Solution: Three vectors are coplanar if there exist scalars \(\lambda ,\mu  \in \mathbb{R}\) using which one vector can be expressed as the linear combination of the other two.

Let us try to find such scalars:

\[2\vec a - \vec b + 3\vec c = \lambda \left( {\vec a + \vec b - 2\vec c} \right) + \mu \left( {\vec a + \vec b - 3\vec c} \right)\]

\[ \Rightarrow  \quad  \left( {2 - \lambda  - \mu } \right)\vec a + \left( { - 1 - \lambda  - \mu } \right)\vec b + \left( {3 + 2\lambda  + 3\mu } \right)\vec c = \vec 0\]

Since \(\vec a,\vec b,\vec c,\) are non-coplanar, we must have

\[2 - \lambda  - \mu  = 0\]

\[ - 1 - \lambda  - \mu  = 0\]

\[3 + 2\lambda  + 3\mu  = 0\]

This system, as can be easily verified , does not have a solution for  \(\lambda \,\,{\text{and}}\,\,\mu \).

Thus, we cannot find scalars for which one vector can be expressed as the linear combination of the other two, implying the three vectors must be non-coplanar.

As an additional exercise, show that for three non-coplanar vectors \(\vec a,\,\,\vec b\,\,{\text{and}}\,\,\vec c\) , the vectors  \(\vec a - \,2\vec b\, + 3\vec c,\,\,\vec a - 3\vec b + 5\vec c\,\,\,and\,\,\, - 2\vec a - \,3\vec b\, - 4\vec c\)  are coplanar.

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